Question

the data shows that the bone lengths have a mean of 45.9 cm and standard deviation of 4.2 cm, the overall heights have a mean of 172.7 cm and standard deviation of 8.14 cm, and the correlation between bone length and height is 0.914.

what is the slope of the LSRL of height on bone length?

Answer 1

least squares regression line?

Answer 2

yes

Answer 3

hmm .... i wonder how we can fit this into the only thing i know for finding the slope of this:\[m=\frac{n\sum xy-\sum x\sum y}{n\sum xx-\sum x\sum x }\]

Answer 4

i know this is an analysis of variance .... type setup

Answer 5

I don't know how to put it in that equation

Answer 6

me either, at the moment, im pretty sure there is some simplified versions that are in the textbook ... all check back in a few minutes

Answer 7

\[\sigma^2=\frac{\sum(x-\mu)^2}{n}=\frac{\sum x^2-2\mu\sum x+\mu^2\sum1}{n}\]
\[\sigma^2=\frac{\sum x^2}{n}-2\mu\frac{\sum x}{n}+\mu^2\frac{\sum1}{n}\]
\[\sigma^2=\frac{\sum x^2}{n}-2\mu\mu+\mu^2\frac{n}{n}\]
\[\sigma^2=\frac{\sum x^2}{n}-\mu^2\]
\[\sigma^2=\frac{\sum x^2}{n}-\frac{\sum x}{n}\frac{\sum x}{n}\]
\[\sigma^2=\frac{\sum x^2-\sum x\sum x}{n}\]
which is prolly how we get to the parts in the slope setup

Answer 8

im confusing r for slope ... but they are very similar

Answer 9

\[m=r\frac{s_y}{s_x}\]
where the s parts are the standard deviations, and r is the linear correlation coeff

Answer 10

any recollection from your material how to find r with the given information?

Answer 11

no I don't have a book or anything

Answer 12

"and the correlation between bone length and height is 0.914"

r = 0.914 , simply read right over that :/

m = 0.914(8.14)/4.2

assuming bone is along the x axis ...